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Question

a 60kg skier starts from rest at the top of a 65m high slope. A) If frictional force do -10.5 J work how fast is she going at the bottom. B) Now moving horizontally the skier crosses a patch of soft snow where the Uk =.20. If the patch is 82 mwide and the asverage force of air on the ski is 160N how fast is she going after crossing the patch. C) the skier hits a snowdrift and penetrates 2.5 m into it before stopping. what is the average force exerted on her by the snow drift

Answer 1

Part A can be solved using energy equations
the initial kinetic energy is 0.
The loss in potential energy is
m*g*h
=60*9.81*65
=38259 J

The loss due to friction is 10.5 J
(Are you sure that's not 10.5 kJ?)

the kinetic at the bottom is
.5*m*v^2
or 30*v^2
=38259-10.5
v=sqrt((38259-10.5)/30)
=35.7 m/s

(if the friction was 10.5 kJ, the v=30.42)

B) Similarly
The work due to air resistance is
82*160
=-13120

the work due to friction is
=-m*g*µ*d
=-60*9.81*.2*82
=-9653

the starting energy was (using 10.5 J)
=.5*m*v^2
=30*35.7^2
=38235

the kinetic at the end of the 82 m patch is
30*v^2=38235-9653-13120
v=sqrt((38235-9653-13120)/30)
=22.7 m/s

(If 10.5 kJ, then v = 12.9 m/s)

C
Since she stopped in 2.5 m, the force was
F*2.5=30*22.7^2
F=7729 N

(if 10.5 kJ, F=2496 N)

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